Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\displaystyle I_n=\int_0^\infty2^{-x}(1+x)^ndx$

By integral by parts, I can get $I_n=\frac{1}{\ln 2}+\frac{n}{\ln 2}I_{n-1}$ where $I_0=\frac{1}{\ln 2}$. How to proceed to conclude an explicit expression of $I_n=\sum\limits_{m=0}^n\cdots$? Thank you.

share|improve this question

2 Answers 2

up vote 8 down vote accepted

First note that $$\int_0^{\infty} 2^{-x} x^m dx = \int_0^{\infty} e^{-x \log2} x^m dx = \dfrac{\Gamma(m+1)}{(\log 2)^{m+1}}$$ Now $$\int_0^{\infty} 2^{-x} (1+x)^n dx = \sum_{k=0}^{n} \dbinom{n}k \int_0^{\infty} 2^{-x} x^k dx = \sum_{k=0}^{n} \dbinom{n}k \dfrac{\Gamma(k+1)}{(\log 2)^{k+1}} = \sum_{k=0}^n \dfrac{P(n,k)}{\log(2)^{k+1}}$$ where $P(n,k) = n \cdot (n-1) \cdot (n-2) \cdots (n-k+1)$

share|improve this answer
    
Cool! Thank you! . –  Zhou Heng Jan 12 '13 at 7:36

Iterate. Since $I_n = \frac{1}{\ln 2}(1 + n I_{n-1})$, then $I_{n-1} = \frac{1}{\ln 2}[1 + (n-1) I_{n-2}]$, and so-on. Repeating down to $I_0$ yields

$$ I_n = \frac{1}{\ln 2}(1 + n I_{n-1}) = \frac{1}{\ln 2}(1 + n[\frac{1}{\ln 2}[1 + (n-1) I_{n-2}]]) \\ = \text{...} = \frac{1}{\ln2} + \frac{n}{(\ln2)^2} + \frac{n(n-1)}{(\ln2)^3} + \frac{n(n-1)(n-2)}{(\ln2)^4} + \text{...} = \sum_{k=0}^n \frac{P(n,k)}{(\ln2)^{k+1}}$$

just as Marvis found using the (very slick) Binomial expansion + Gamma function approach above.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.